Abstract
Some of the oldest combinatorial objects, whose study apparently goes back to ancient times, are the Latin squares. To obtain a Latin square, one has to fill the n 2 cells of an (n x n)-square array with the numbers 1,2,…, n so that that every number appears exactly once in every row and in every column. In other words, the rows and columns each represent permutations of the set {1,…, n}. Let us call n the order of the Latin square. Here is the problem we want to discuss. Suppose someone started filling the cells with the numbers {1, 2,…, n}. At some point he stops and asks us to fill in the remaining cells so that we get a Latin square. When is this possible? In order to have a chance at all we must, of course, assume that at the start of our task any element appears at most once in every row and in every column. Let us give this situation a name. We speak of a partial Latin square of order n if some cells of an (n x n)-array are filled with numbers from the set {1,…, n} such that every number appears at most once in every row and column. So the problem is:
When can a partial Latin square be completed to a Latin square of the same order?
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References
T. Evans: Embedding incomplete Latin squares, Amer. Math. Monthly 67(1960), 958–961.
C. C. Lindner: On completing Latin rectangles, Canadian Math. Bulletin 13(1970), 65–68.
H.J. Ryser: A combinatorial theorem with an application to Latin rectangles, Proc. Amer. Math. Soc. 2 (1951), 550–552.
B. Smetaniuk: A new construction on Latin squares I: A proof of the Evans conjecture, Ars Combinatoria 11 (1981), 155–172.
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© 2004 Springer-Verlag Berlin Heidelberg
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Aigner, M., Ziegler, G.M. (2004). Completing Latin squares. In: Proofs from THE BOOK. Springer, Berlin, Heidelberg. https://doi.org/10.1007/978-3-662-05412-3_27
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DOI: https://doi.org/10.1007/978-3-662-05412-3_27
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